Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\frac{2}{3}+\frac{2}{3}+\frac{6}{3}=\frac{10}{3}\)
b,\(\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=\frac{6}{4}+\frac{3}{2}=\frac{3}{2}+\frac{3}{2}=\frac{6}{2}=3\)
\(\frac{12}{18}+\frac{1}{3}+\frac{1}{7}+\frac{2}{8}+\frac{27}{36}+\frac{42}{49}\)
\(=\frac{2}{3}+\frac{1}{3}+\frac{1}{7}+\frac{1}{4}+\frac{3}{4}+\frac{6}{7}\)
\(=\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\left(\frac{1}{4}+\frac{3}{4}\right)\)
\(=1+1+1\)
\(=3\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}...\frac{1}{7x8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)\(-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
b,
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}+\frac{1}{2916}\)
\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}+\frac{1}{972}\)
\(3A-A=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}+\frac{1}{972}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}+\frac{1}{2916}\right)\)
\(2A=\frac{3}{4}-\frac{1}{2916}\)
\(A=\frac{1093}{2916}\)
Tính A rồi so sánh:
\(\frac{1}{4}\) + \(\frac{1}{9}\) + \(\frac{1}{16}\) + \(\frac{1}{25}\) + \(\frac{1}{36}\) = \(\frac{1769}{3600}\)
3600 chia hết cho 6, nên ta chọn 3600 làm mẫu số chung.
3600 : 6 = 600
\(\frac{5}{6}\) = \(\frac{5\times600}{6\times600}\) = \(\frac{3000}{3600}\)
Mà \(\frac{3000}{3600}\) > \(\frac{1769}{3600}\)
Nên: \(\frac{5}{6}\) > A
\(\left(2.8x-32\right):\frac{2}{3}=90\)
\(2.8\cdot x-32=90\cdot\frac{2}{3}\)
\(\frac{14}{5}x-32=60\)
\(\frac{14}{5}x=60+32\)
\(\frac{14}{5}x=92\)
\(x=\frac{230}{7}\)
B , c , d tương tự
1093/2916
A = \(\frac{1093}{2916}\)